<span>30.0 ml of 0.15 m K2CrO4 solution will have more potassium ions.
Let's see the relative number of potassium ions for each solution. Since all the measurements are the same, the real difference is the K2CrO4 will only have 2 potassium ions per molecule while the K3PO4 solution will have 3 potassium ions per molecule.
K2CrO4 solution
30.0 * 0.15 * 2 = 9
K3PO4 solution
25.0 * 0.080 * 3 = 6
Since 9 is greater than 6, the K2CrO4 solution will have more potassium ions.</span>
Answer:
The new temperature will be 2546 K or 2273 °C
Explanation:
Step 1: Data given
The initial temperature = 1000 °C =1273 K
The volume = 20L
The volume increases to 40 L
Step 2: Calculate the new temperature
V1/T1 = V2/T2
⇒with V1 = the initial volume = 20L
⇒with T1 = the initial temperature = 1273 K
⇒with V2 = the increased volume = 40L
⇒with T2 = the new temperature = TO BE DETERMINED
20L/ 1273 K = 40L / T2
T2 = 40L / (20L/1273K)
T2 = 2546 K
The new temperature will be 2546 K
This is 2546-273 = 2273 °C
Since the volume is doubled, the temperature is doubled as well
30kg think of it like this imagine you are moving to your college dorm would you have you carry the box if weights or will you carry your books instead
Answer:
The percent isotopic abundance of Ir-193 is 60.85 %
The percent isotopic abundance of Ir-191 is 39.15 %
Explanation:
we know there are two naturally occurring isotopes of iridium, Ir-191 and Ir-193
First of all we will set the fraction for both isotopes
X for the isotopes having mass 193
1-x for isotopes having mass 191
The average atomic mass is 192.217
we will use the following equation,
193x + 191(1-x) = 192.217
193x + 191 - 191x = 192.217
193x- 191x = 192.217 - 191
2x = 1.217
x= 1.217/2
x= 0.6085
0.6085 × 100 = 60.85 %
60.85% is abundance of Ir-193 because we solve the fraction x.
now we will calculate the abundance of Ir-191.
(1-x)
1-0.6085 =0.3915
0.3915× 100= 39.15 %
